Date: 02/14/98 at 21:06:02 From: Jamie Hanson Subject: Indirect Proofs Prove that sqrt 2 is an irrational number. I have no ideas. Can you help?
Date: 02/17/98 at 19:14:02 From: Doctor Wolf Subject: Re: Indirect Proofs Hi Jamie, Let's say that sqrt(2) was rational; that is, it could be written as a fraction a/b in REDUCED form where both a and b are positive integers. Reduced form would imply that the integers a and b could not both be even, otherwise we could reduce the fraction even further by dividing both a and b by 2 (example: 8/12 = 4/6 = 2/3). Since a/b = sqrt(2) we would have a^2 2 --------- = --- (squaring both b^2 1 sides of equation) It gets a little tricky here ... so let's go slowly! There is a property of numbers I will need (not too hard to prove): Only even integers will produce even integers when squared. Back to the problem. If I cross multiply in the equation above, I'll have a^2 = 2b^2. Well, 2b^2 is certainly an even integer (since it has 2 as a factor), and so both a^2 AND a (using number property) must be even. Okay ... so we've shown a is an even integer ... now what? What if we could show that b is also even ... a contradiction! Since a is an even integer, we can write it as 2c where c is just a divided by 2 (for example: 10 = 2*5). Now let's rewrite our original equation replacing a by 2c: (2c)^2 2 4c^2 2 ------ = - or ---- = --- or (after cross multiplying) b^2 1 b^2 1 4c^2 = 2b^2. After dividing both sides by 2, 2c^2 = b^2. What does this say (odd or even integer) about 2c^2, b^2, and b? And why is this a contradiction of our original assumption about a/b? Can you take it from here Jamie ...(Remember to use number property.) P.S.: There are lots of questions like yours and their solutions in the Dr. Math Archives ... you might want to check them out. -Doctor Wolf, The Math Forum Check out our Web site http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum